Question

review & refresh
in exercises 20 and 21, solve the equation using any method. explain your choice of method.

1. $4x^{2}-87 = 109$
2. $3x^{2}-2x - 7 = 0$
3. does the table represent a linear or nonlinear function? explain.

x	2	4	6	8	10
y	$\frac{1}{2}$	1	2	4	8

in exercises 23 and 24, find the angle measure.

1. $angle1$ is a complement of $angle4$, and $mangle1 = 33^{circ}$. find $mangle4$.
2. $angle3$ is a supplement of $angle2$, and $mangle2 = 147^{circ}$. find $mangle3$.

Explanation:

Exercise 20

Step1: Isolate the $x^{2}$ - term

Add 87 to both sides of the equation $4x^{2}-87 = 109$.
$4x^{2}=109 + 87=196$

Step2: Solve for $x^{2}$

Divide both sides of the equation $4x^{2}=196$ by 4.
$x^{2}=\frac{196}{4}=49$

Step3: Solve for $x$

Take the square - root of both sides.
$x=\pm\sqrt{49}=\pm7$

We use the quadratic formula for a quadratic equation of the form $ax^{2}+bx + c = 0$, which is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For the equation $3x^{2}-2x - 7 = 0$, we have $a = 3$, $b=-2$, and $c=-7$.

Step1: Calculate the discriminant $\Delta=b^{2}-4ac$

$\Delta=(-2)^{2}-4\times3\times(-7)=4 + 84 = 88$

Step2: Find the values of $x$

$x=\frac{-(-2)\pm\sqrt{88}}{2\times3}=\frac{2\pm2\sqrt{22}}{6}=\frac{1\pm\sqrt{22}}{3}$

For a linear function, the rate of change (slope) is constant. Calculate the slope between consecutive points.
The slope $m$ between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}$.
Between $(2,\frac{1}{2})$ and $(4,1)$: $m_1=\frac{1-\frac{1}{2}}{4 - 2}=\frac{\frac{1}{2}}{2}=\frac{1}{4}$
Between $(4,1)$ and $(6,2)$: $m_2=\frac{2 - 1}{6 - 4}=\frac{1}{2}$
Since the slopes are not constant, the function is nonlinear.

Answer:

$x = 7$ or $x=-7$

Exercise 21